You have found the following ages (in years) of 6 porcupines. The porcupines are randomly selected from the 36 porcupines at your local zoo: $ 4,\enspace 15,\enspace 14,\enspace 6,\enspace 2,\enspace 10$ Based on your sample, what is the average age of the porcupines? What is the variance? You may round your answers to the nearest tenth.
Because we only have data for a small sample of the 36 porcupines, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{20.25} + {42.25} + {30.25} + {6.25} + {42.25} + {2.25}} {{6 - 1}} $ $ {s^2} = \dfrac{{143.5}}{{5}} = {28.7\text{ years}^2} $ We can estimate that the average porcupine at the zoo is 8.5 years old. There is a variance of 28.7 years $^2$.